Consider the oscillation Duffing equation
$$ F(x,\mu, \beta)=x''+x+\mu x^{3}-\beta\sin t = 0,\quad \beta > 0, \mu\neq0, $$
(13)
where \(F: \mathcal{C}^{2}[0,2\pi]\times\mathbb{R}\times \mathbb{R}\rightarrow\mathcal{C}[0,2\pi]\) and \(x(0)=x(2\pi)=0\).
Note that for \(\mu=0\), \(\beta> 0\), solutions of the above equation do not exist, but for \(\beta=0\), there exists zero solution. We are interested in nontrivial solutions of this equation of course. The following theorem holds.
Theorem 6
For sufficiently small
\(\beta> 0\), the Duffing equation (1) has nontrivial solution tangent to elements from the 4-kernel of the form
$$\begin{aligned}& x(\beta,t) = \frac{\beta^{\frac{1}{4}}}{\overline {\beta}^{\frac{1}{4}}} \biggl(\sqrt[3]{\frac{4\overline{\beta}}{3 \overline{\mu}}}\sin t \biggr)+ o\bigl(\beta^{\frac{1}{4}}\bigr), \end{aligned}$$
(14)
$$\begin{aligned}& \mu(\beta) = \frac{\beta^{\frac{1}{4}}}{\overline {\beta}^{\frac{1}{4}}} \overline{\mu} + o\bigl(\beta^{\frac{1}{4}} \bigr), \end{aligned}$$
(15)
where
μ̅, β̅
are fixed numbers for which the following conditions hold: \(3\sqrt[3]{\frac{4\overline{\beta}}{3 \overline{\mu}}}+ \vert \overline{\mu} \vert + \vert \overline{\beta}^{\frac{1}{4}} \vert =1\), \(\overline{\mu}=\overline{\mu}(\overline{\beta}^{\frac{1}{4}})\), \(\overline{\beta}=\overline{\alpha}^{4}\)
and
\(0<\overline{\beta}^{\frac{1}{4}}<1\).
Proof
In order to apply the theory of p-regularity, let us put \(\beta^{\frac{1}{4}} = \alpha\) and go to an equivalent equation of the form
$$ \overline{F}(x,\mu, \alpha)=x''+x+\mu x^{3}- \alpha^{4} \sin t = 0. $$
(16)
Exactly now we show that the mapping F̅ is 4-regular, and we will describe solutions. Of course, \(x^{\ast}=(0,0,0)\) ais the trivial solution of the above equation. We have
$$ \begin{aligned}[b] \overline{F}'(x,\mu, \alpha)&= \bigl(\overline{F}'_{x}(x, \mu, \alpha),\overline{F}'_{\mu}(x,\mu, \alpha), \overline{F}'_{\alpha}(x,\mu, \alpha)\bigr)\\ &= \biggl( \frac{d^{2}}{dt^{2}}+1+3\mu x^{2},x^{3},-4\alpha^{3} \sin t \biggr) \end{aligned} $$
(17)
and
$$ \overline{F}'(0,0,0)=\bigl(\overline{F}'_{x}(0,0,0), \overline{F}'_{\mu }(0,0,0),\overline{F}'_{\alpha}(0,0,0) \bigr)= \biggl(\frac{d^{2}}{dt^{2}}+1,0,0 \biggr). $$
Note that \(\operatorname{Ker} \overline{F}'_{x}(0,0,0)=\{x\in \mathcal{C}^{2}[0,2\pi]:\frac{d^{2}x}{dt^{2}}+x=0\}\). The general solutions of the equation \(x''+x=0\) are \(x(t)=c_{1}\cos t+c_{2}\sin t\). Taking into account the boundary conditions, we obtain \(c_{1}=0\), \(x(t)=c_{2}\sin t\) and \(\operatorname{Ker} \overline{F}'_{x}(0,0,0)=\operatorname{span}\{\sin t\}\).
The image of the operator \(\overline{F}'_{x}(0,0,0)\) is defined as follows:
$$\begin{aligned} \operatorname{Im} \overline{F}'_{x}(0,0,0) =&\bigl\{ y\in \mathcal{C}[0,2\pi ]:\exists x\in\mathcal{C}^{2}[0,2\pi] \overline{F}'_{x}(0,0,0)x=y, x(0)=x(2\pi)=0\bigr\} \\ =&\bigl\{ y\in\mathcal{C}[0,2\pi]:\exists x\in\mathcal{C}^{2}[0,2\pi] x''+x=y, x(0)=x(2\pi)=0\bigr\} . \end{aligned}$$
The general solution of the equation \(x''+x=y\) has the form
$$ x(t)=c_{1}\cos t+c_{2}\sin t-\cos t \int_{0}^{t}y(\tau)\sin \tau \,d\tau+\sin t \int_{0}^{t}y(\tau)\cos\tau \,d\tau. $$
In view of the boundary condition, we obtain \(c_{1}=0\) i \(\int_{0}^{2\pi}y(\tau)\sin\tau \,d\tau=0\), then
$$\begin{aligned} \operatorname{Im} \overline{F}'_{x}(0,0,0)&=\biggl\{ y\in \mathcal{C}[0,2\pi]: \int _{0}^{2\pi}y(\tau)\sin\tau \,d\tau=0\biggr\} \\ &=\bigl\{ y\in\mathcal{C}[0,2\pi]:\langle y,\sin t \rangle=0,\sin t\in \operatorname{Ker} \overline{F}'_{x}(0,0,0)\bigr\} . \end{aligned} $$
One can easily show that the boundary value problem \(x''+x=\sin t\), \(x(0)=x(2\pi)=0\) does not have a solution. This implies that the operator \(\overline{F}'_{x}(0,0,0)\) is not surjective and \(\operatorname{Im} \overline{F}'_{x}(0,0,0)\neq\mathcal{C}[0,2\pi]=Y\). Then \(Y=Y_{1}\oplus Z_{2}\), where \(Y_{1}=\operatorname{Im} \overline{F}'_{x}(0,0,0)\), \(Z_{2}=Y_{1}^{\bot}\).
The projector \(P_{Z_{2}}:Y\rightarrow Z_{2}\) can be described as
$$ P_{Z_{2}}y=\frac{1}{\pi}\sin t\langle y,\sin t\rangle= \frac{1}{\pi}\sin t \int_{0}^{2\pi}y(\tau)\sin\tau \,d\tau,\quad y\in Y. $$
This implies that
$$\begin{aligned} Y_{2} =& \operatorname{span}\bigl(\operatorname{Im} P_{Z_{2}} \overline {F}''(0,0,0)[\cdot]^{2}\bigr) \\ =& \operatorname{span}\biggl\{ y(t)\in Y:y(t)=\frac{1}{\pi}\sin t \int_{0}^{2\pi}\overline{F}''(0,0,0) \bigl[z(\tau ) \bigr]^{2}\sin\tau \,d\tau, z(\tau)\in \mathcal{C}^{2}[0,2\pi] \biggr\} \\ \subseteq& Z_{2}. \end{aligned}$$
Let us evaluate the second derivative of the mapping F̅
$$ \overline{F}''(x,\mu,\alpha)=\bigl(\bigl(6 \mu x,3x^{2},0\bigr),\bigl(3x^{2},0,0\bigr),\bigl(0,0,-12 \alpha^{2} \sin t\bigr)\bigr). $$
From this we obtain
$$ \overline{F}''(0,0,0)=0 $$
and \(Y_{2}=\{0\}\), \(Z_{3}=(Y_{1} \oplus\{0\})^{\bot}\). We continue
$$\begin{aligned}& P_{Z_{3}}y=\frac{1}{\pi}\sin t\langle y,\sin t\rangle = \frac{1}{\pi}\sin t \int_{0}^{2\pi}y(\tau)\sin\tau \,d\tau,\quad y\in Y, \\& \begin{aligned} Y_{3} &= \operatorname{span}\bigl(\operatorname{Im} P_{Z_{3}} \overline {F}'''(0,0,0)[ \cdot]^{3}\bigr) \\ &=\operatorname{span}\biggl\{ y(t)\in Y:y(t) = \frac{1}{\pi}\sin t \int_{0}^{2\pi}\overline{F}'''(0,0,0) \bigl[z(\tau ) \bigr]^{3}\sin\tau \,d\tau, z(\tau)\in \mathcal{C}^{2}[0,2\pi] \biggr\} \\ &\subseteq Z_{3}. \end{aligned} \end{aligned}$$
Let us evaluate the third derivative of the mapping F̅:
$$\begin{aligned} \overline{F}'''(x,\mu,\alpha) =&\bigl( \bigl((6\mu,6x,0) (6x,0,0),(0,0,0)\bigr),\bigl((6x,0,0),(0,0,0),(0,0,0)\bigr), \\ &\bigl((0,0,0),(0,0,0),(0,0,-24\alpha\sin t)\bigr)\bigr). \end{aligned}$$
From this
$$ \overline{F}'''(0,0,0)=0 $$
and \(Y_{3}=\{0\}\) i \(Y= Y_{1}\oplus\{0\}\oplus\{0\}\oplus Z_{4}\), \(Z_{4}=(Y_{1} \oplus\{0\}\oplus\{0\})^{\bot}\). Next we have
$$ P_{Z_{4}}y=\frac{1}{\pi}\sin t\langle y,\sin t \rangle= \frac{1}{\pi}\sin t \int_{0}^{2\pi}y(\tau)\sin\tau \,d\tau,\quad y\in Y $$
and
$$\begin{aligned} Y_{4} =& \operatorname{span}\bigl(\operatorname{Im} P_{Z_{4}} \overline {F}^{(4)}(0,0,0)[\cdot]^{4}\bigr) \\ =&\operatorname{span} \biggl\{ y(t)\in Y:y(t)=\frac{1}{\pi}\sin t \int_{0}^{2\pi}\overline {F}^{(4)}(0,0,0) \bigl[z(\tau) \bigr]^{4}\sin\tau \,d\tau, z(\tau)\in \mathcal{C}^{2}[0,2\pi] \biggr\} . \end{aligned}$$
Let us evaluate the fourth derivative:
$$\begin{aligned}& \overline{F}^{(4)}(x,\mu,\alpha) \\& \quad = \bigl(\bigl(\bigl((0,6,0),(6,0,0),(0,0,0)\bigr),\bigl((6,0,0),(0,0,0),(0,0,0) \bigr), \\& \qquad \bigl((0,0,0),(0,0,0),(0,0,0)\bigr)\bigr),\bigl(\bigl((6,0,0),(0,0,0),(0,0,0) \bigr), \\& \qquad \bigl((0,0,0),(0,0,0),(0,0,0)\bigr),\bigl((0,0,0),(0,0,0),(0,0,0)\bigr)\bigr), \\& \qquad \bigl(\bigl((0,0,0),(0,0,0),(0,0,0)\bigr),\bigl((0,0,0),(0,0,0),(0,0,0)\bigr), \\& \qquad \bigl((0,0,0),(0,0,0),(0,0,-24\sin t)\bigr)\bigr)\bigr)\\& \quad =\overline{F}^{(4)}(0,0,0). \end{aligned}$$
Note that \(\overline{F}^{(5)}(x,\mu,\alpha)=0\), \(Z_{4}=Y_{4}\) and \(P_{Z_{4}}y = \Pi_{Y_{4}}y\). Therefore we will examine 4-regularity.
Now let us take \(z(\tau)=(h_{x},h_{\mu},h_{\alpha})\). For such a defined vector \(z(\tau)\), the following relation holds:
$$\begin{aligned}& \overline{F}^{(4)}(0,0,0)\bigl[z(\tau)\bigr]^{3}= \bigl(18h_{x}^{2}h_{\mu },6h_{x}^{3},-24 \sin \tau h_{\alpha}^{3}\bigr), \end{aligned}$$
(18)
$$\begin{aligned}& \overline{F}^{(4)}(0,0,0)\bigl[z(\tau)\bigr]^{4}=24 \bigl(h_{x}^{3}h_{\mu}-\sin \tau h_{\alpha}^{4}\bigr), \end{aligned}$$
(19)
and we can describe the subspace
$$\begin{aligned} Y_{4} =& \operatorname{span}\biggl\{ y(t)\in Y: y(t)=\frac{24}{\pi}\sin t \int _{0}^{2\pi}\bigl(h_{x}^{3}h_{\mu}- \sin\tau h_{\alpha}^{4}\bigr)\sin\tau \,d\tau\biggr\} \\ =& \operatorname{span}\{\sin t\}=\operatorname{Ker} \overline{F}'_{x}(0,0,0), \end{aligned}$$
4-factor operator
$$\begin{aligned}& \forall u=(h_{u},h_{\lambda},h_{\beta})\in C^{2}[0,2\pi]\times \mathbb{R} \times \mathbb{R} \\& \begin{aligned} \Psi_{4}(h)[u]&=\Psi_{4}\bigl((0,0,0),(h_{x},h_{\mu},h_{\alpha }) \bigr) (h_{u},h_{\lambda},h_{\beta}) \\ & = \frac{d^{2}h_{u}}{dt^{2}}+h_{u} + \frac{1}{\pi}\sin t \int_{0}^{2\pi}\bigl(18h_{x}^{2}h_{\mu}h_{u} + 6h_{x}^{3}h_{\lambda}- 24 \sin\tau h_{\alpha}^{3}h_{\beta}\bigr)\sin\tau \,d\tau \end{aligned} \end{aligned}$$
and the 4-kernel of 4-factor operator \(\Psi_{4}(h)\)
$$\begin{aligned} \operatorname{Ker}^{4} \Psi_{4}(h)&=\biggl\{ h=(h_{x},h_{\mu},h_{\alpha}) \in C^{2}[0,2\pi]\times\mathbb{R} \times \mathbb{R}: \\ &\quad \frac{d^{2}h_{x}}{dt^{2}}+h_{x} + \frac{24}{\pi}\sin t \int_{0}^{2\pi}\bigl(h_{x}^{3}h_{\mu}- \sin\tau h_{\alpha}^{4}\bigr)\sin \tau \,d\tau=0\biggr\} . \end{aligned}$$
Taking into account the equations
$$ \int_{0}^{2\pi}\sin^{4}\tau \,d\tau= \frac{3}{4}\pi, \qquad \int_{0}^{2\pi}\sin^{2}\tau \,d\tau=\pi $$
and the fact that \(h_{x}=z\sin\tau\), since \(h_{x}\in\operatorname{Ker} \overline{F}'_{x}(0,0,0)\), we solve the following equation of unknowns z, \(h_{\mu}\), \(h_{\alpha}\):
$$ \frac{24}{\pi}\sin t \biggl[z^{3}h_{\mu} \int_{0}^{2\pi}\sin^{4} \tau \,d \tau- h_{\alpha}^{4} \int_{0}^{2\pi}\sin^{2} \tau \,d\tau \biggr]=0. $$
(20)
From here \(\frac{3}{4}z^{3}h_{\mu}-h_{\alpha}^{4}\)=0 and \(z = \sqrt[3]{\frac{4h_{\alpha}^{4}}{3h_{\mu}}}\) for \(h_{\mu} \neq 0\). We assume here that \(\Vert h \Vert = \Vert (h_{x},h_{\mu}, h_{\alpha}) \Vert =1\) (the vector h lies on a ball of radius 1), where we define the norm \(\Vert \cdot \Vert \) in the space \(\mathcal{C}^{2}[0,2\pi]\) as follows:
$$ \forall x\in\mathcal{C}^{2}[0,2\pi]\quad \Vert x \Vert = \sup _{t\in [0,2\pi]} \bigl\vert x(t) \bigr\vert +\sup_{t\in[0,2\pi]} \bigl\vert x'(t) \bigr\vert +\sup_{t\in [0,2\pi]} \bigl\vert x''(t) \bigr\vert , $$
(21)
while the norm in the space \(\mathbb{R}\times\mathbb{R}\) as \(\Vert (y,z) \Vert = \vert y \vert + \vert z \vert \). We can accept the norm in the space \(\mathcal{C}^{2}[0,2\pi]\times\mathbb{R}\times\mathbb{R}\) as standard, i.e.,
$$ \bigl\Vert (x,y,z) \bigr\Vert = \Vert x \Vert _{\mathcal{C}^{2}[0,2\pi]}+ \bigl\Vert (y,z) \bigr\Vert _{\mathbb{R}\times \mathbb{R}}. $$
(22)
Then we describe the space \(\operatorname{Ker}^{4} \Psi_{4}(h)\) clearly as follows:
$$ \operatorname{Ker}^{4} \Psi_{4}(h)= \biggl\{ \biggl( \sqrt[3]{ \frac{4\alpha^{4}}{3 \mu}}\sin t,\mu, \alpha\biggr) \biggr\} \cup\bigl\{ (c\sin t, 0,0) \bigr\} \cup \bigl\{ (0,\mu,0)\bigr\} , \bigl\Vert (h_{x},\mu,\alpha) \bigr\Vert =1, $$
where \(h_{x}\) equals \(\sqrt[3]{\frac{4\alpha^{4}}{3 \mu}}\sin t\), \(c\sin t\) for \(c\in\mathbb{R}\) or 0.
At the end we will examine the surjectivity of 4-factor operator on the 4-kernel. Note that vectors \((c\sin t,0,0)\) and \((0,\mu,0)\) are solutions of the Duffing equation, then we are interested in vectors of the form \(H=(\sqrt[3]{\frac{4\alpha^{4}}{3 \mu}}\sin t,\mu, \alpha)\), where \(3\sqrt[3]{\frac{4\alpha^{4}}{3 \mu}}+ \vert \mu \vert + \vert \alpha \vert =1\). So we will verify whether
$$ \forall y\in\mathcal{C}[0,2\pi] \exists u=[h_{u},h_{\lambda},h_{\beta}] \in\mathcal{C}^{2}[0,2\pi]\times \mathbb{R}\times\mathbb{R}\quad \Psi_{4}(H)[u]=y. $$
We have
$$\begin{aligned} \Psi_{4}(H)[u] =&\frac{d^{2}h_{u}}{dt^{2}}+h_{u} + \frac{1}{\pi}\sin t \int_{0}^{2\pi} \biggl(18 \biggl(\sqrt[3]{ \frac{4\alpha^{4}}{3 \mu}} \biggr)^{2}\sin^{2} \tau\mu h_{u}\\ &{} + 8\frac{\alpha ^{4}}{\mu}\sin^{3} \tau h_{\lambda}- 24 \sin\tau \alpha^{3}h_{\beta} \biggr)\sin\tau \,d\tau\\ =&y. \end{aligned}$$
Using now Lemma 10, p.6 from [8], i.e., putting \(y= y_{4}=a\sin t\in Y_{4}\), we find the element \(h_{u}= b \sin t \in \operatorname{Ker} \overline{F}'_{x}(0,0,0)\). Namely, we obtain
$$ \frac{1}{\pi}\sin t \int_{0}^{2\pi} \biggl(18 \biggl(\sqrt[3]{ \frac{4\alpha^{4}}{3 \mu}} \biggr)^{2} \mu b \sin^{4} \tau+ 8 \frac{\alpha ^{4}}{\mu}\sin^{4} \tau h_{\lambda}- 24 \alpha^{3}h_{\beta}\sin^{2} \tau \biggr) \,d \tau= a \sin t. $$
This implies \(b= \frac{1}{9\sqrt[3]{6\alpha^{8}\mu}} (a+24\alpha^{3}h_{\beta} - \frac{6 \alpha^{4} }{ \mu} h_{ \lambda } )\) and
$$ h_{u}= \frac{1}{9\sqrt[3]{6\alpha^{8}\mu}} \biggl(a+24\alpha^{3}h_{\beta} - \frac{6 \alpha^{4} }{ \mu} h_{ \lambda } \biggr)\sin t. $$
The solutions of the above equation exist, hence 4-factor operator is a surjection on the element H.
Summing up, we conclude from the generalized Lyusternic theorem: for \(x^{\ast}=(0,0,0)\), the tangent cone \(T_{x\ast}M\) to the solutions set M coincides with the kernel of the fourth derivative of the mapping \(\Psi_{4}\), i.e., with the set \(H_{4}(x^{\ast})= \operatorname{Ker} ^{4}\Psi_{4}(H)\). Thus there exist nontrivial solutions of the Duffing equation (1), and we can write them in the form \(x(\varepsilon, t)=x^{\ast}+\varepsilon H+r(\varepsilon)\), with \(\Vert r(\varepsilon) \Vert =o(\varepsilon)\), for \(\varepsilon\in(0, \delta)\), where \(\delta> 0\) is sufficiently small. Let β̅, μ̅ be fixed numbers such that the equation \(3\sqrt[3]{\frac{4\overline{\beta}}{3 \overline{\mu}}}+ \vert \overline{\mu} \vert + \vert \overline{\beta}^{\frac{1}{4}} \vert =1\) holds, where \(\overline{\mu}=\overline{\mu}(\overline{\beta}^{\frac{1}{4}})\), \(\overline{\beta}=\overline{\alpha}^{4}\) and \(0<\overline{\beta}^{\frac{1}{4}}<1\). Putting \(x^{\ast}=(0,0,0)\), and \(\varepsilon=\frac{\beta^{\frac{1}{4}}}{\overline {\beta}^{\frac{1}{4}}}\), we get formulas (14), (15), and this finishes the proof of the theorem. □